Answer:
1. 0.97 V
2. 
Explanation:
In this case, we can start with the <u>half-reactions</u>:
With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:
I hope it helps!
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Answer
is: volume is 20 mL.<span>
c</span>₁(CH₃COOH) = 2,5 M.<span>
c</span>₂(CH₃COOH) = 0,5 M.<span>
V</span>₂(CH₃COOH) = 100 mL.<span>
V</span>₁(CH₃COOH) = ?<span>
c</span>₁(CH₃COOH) · V₁(CH₃COOH)
= c₂(CH₃COOH) · V₂(CH₃COOH).<span>
2,5 M · V</span>₁(CH₃COOH)
= 0,5 M · 100 mL.<span>
V</span>₁(CH₃COOH) = 0,5 M · 100 mL ÷ 2,5 M.<span>
V</span>₁(CH₃COOH) = 20 mL ÷ 1000 mL/L =0,02 L.
Answer:
look in the explanation part
Explanation:
In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.
Answer:
the radius is half the diameter
